Current Affairs updated daily
KalviGavaniKalviGavani

Samacheer Kalvi 7th Maths (Term 2) Measurements Exercise Solutions

Complete Samacheer Kalvi 7th Maths Term 2 Chapter 2 Measurements exercise solutions — circles, circumference, area, perimeter, semicircles, combined shapes solved step-by-step.

Updated 12 June 202617 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

Samacheer Kalvi 7th Maths (Term 2) — Measurements: Exercise Solutions

How to Use This Page

1. Attempt every question on your own first — work through the problem with pen and paper before scrolling to the solution. 2. Compare your method, not just your answer — focus on how each step follows from the one before. 3. Note the formula used at the start of each solution so you can memorise the key relationships. 4. Re-work any question you got wrong immediately after reading the solution; this cures recurring mistakes faster than passive reading. 5. Use the Key Concepts section as a quick-revision checklist before exams.

Key Concepts (Quick Recap)

  • Circumference of a circle: C = 2πr (when radius is given) or C = πd (when diameter is given), where π ≈ 22/7 ≈ 3.14.
  • Area of a circle: A = πr², where r is the radius.
  • Perimeter of a semicircular disc: The curved part is half the circumference (πr) and the straight edge is the diameter (2r), so Perimeter = πr + 2r.
  • Area between two concentric circles (ring/path): π(R² − r²), where R is the outer radius and r is the inner radius.
  • Revolution problems: Distance covered in one revolution = circumference of the wheel. Number of revolutions = Total distance ÷ Circumference.
  • Ratio of circumferences = ratio of radii: C₁/C₂ = r₁/r₂ (because C = 2πr, so the 2π cancels).
  • Ratio of areas = ratio of squares of radii: A₁/A₂ = r₁²/r₂².
  • Unit conversions: 1 m = 100 cm, 1 km = 1000 m, 1 dm = 10 cm. Always convert all quantities to the same unit before calculating.

Exercise 2.1 — Circumference and Perimeter of Circles

Q1. A circular disc of radius 28 cm is divided into two equal parts. Find the perimeter of each semicircular shape and the perimeter of the circular disc.

Solution
GivenRadius of circular disc, r = 28 cm.

Step 1 — Find the circumference of the full circular disc.

Circumference = 2πr

= 2 × (22/7) × 28

= 2 × 22 × 4

= 176 cm

Step 2 — Find the perimeter of each semicircular shape.

When a circular disc is divided into two equal semicircles, each semicircular piece has:

  • A curved part (half the circumference) = πr
  • A straight edge (the diameter) = 2r

Perimeter of semicircular disc = πr + 2r

= (22/7) × 28 + 2 × 28

= 22 × 4 + 56

= 88 + 56

= 144 cm

AnswerPerimeter of circular disc = 176 cm; Perimeter of each semicircular shape = 144 cm.

Q2. A gardener wants to fence a circular garden of diameter 14 m with 3 rounds of fence. Find the length of rope needed and the cost at ₹4 per metre.

Solution
GivenDiameter d = 14 m, Number of rounds = 3, Cost per metre = ₹4.

Step 1 — Find the circumference of the garden.

C = πd

= (22/7) × 14

= 22 × 2

= 44 m

Step 2 — Find the total length of rope for 3 rounds.

Length of rope = 3 × Circumference

= 3 × 44

= 132 m

Step 3 — Find the total cost.

Cost = ₹4 × 132 = ₹528

AnswerLength of rope = 132 m; Total cost = ₹528.

Q3. Latha wants lace on the edge of a circular table cover of diameter 3 m. Find the length of lace needed and the cost at ₹30 per metre (use π = 3.15).

Solution
GivenDiameter d = 3 m, Cost per metre = ₹30, π = 3.15.

Step 1 — Find the circumference (length of lace needed).

C = πd

= 3.15 × 3

= 9.45 m

Step 2 — Find the total cost.

Cost = ₹30 × 9.45

= ₹(30 × 9 + 30 × 0.45)

= ₹270 + ₹13.50

= ₹283.50

AnswerLength of lace = 9.45 m; Total cost = ₹283.50.

Q4. Find the circumference of a circle with radius 21 cm.

Solution
GivenRadius r = 21 cm.

C = 2πr

= 2 × (22/7) × 21

= 2 × 22 × 3

= 132 cm

AnswerCircumference = 132 cm.

Q5. Find the circumference of a circle with diameter 56 cm.

Solution
GivenDiameter d = 56 cm.

C = πd

= (22/7) × 56

= 22 × 8

= 176 cm

AnswerCircumference = 176 cm.

Q6. The ratio of the circumferences of two circles is 3 : 5. Find the ratio of their radii.

Solution
GivenC₁ : C₂ = 3 : 5.

Step 1 — Write the formula relationship.

C₁ = 2πr₁ and C₂ = 2πr₂

Step 2 — Form the ratio.

C₁ / C₂ = 2πr₁ / 2πr₂

The factor 2π cancels out:

r₁ / r₂ = C₁ / C₂ = 3 / 5

AnswerThe ratio of their radii = 3 : 5.

This is because circumference is directly proportional to radius — if circumferences are in ratio 3 : 5, radii are also in ratio 3 : 5.

Q7. A wheel has diameter 84 cm. How many revolutions does it make to cover 1.056 km?

Solution
GivenDiameter d = 84 cm, Total distance = 1.056 km.

Step 1 — Find the circumference of the wheel.

C = πd

= (22/7) × 84

= 22 × 12

= 264 cm

Step 2 — Convert the total distance to centimetres.

1.056 km = 1.056 × 1000 m = 1056 m

1056 m = 1056 × 100 cm = 1,05,600 cm

Step 3 — Alternatively, convert circumference to metres.

C = 264 cm = 2.64 m

Step 4 — Find the number of revolutions.

Number of revolutions = Total distance / Distance per revolution

Using metres: Number of revolutions = 1056 / 2.64

= 105600 / 264

Let us simplify: 264 × 4 = 1056, so 264 × 400 = 1,05,600

Number of revolutions = 400

AnswerThe wheel makes 400 revolutions to cover 1.056 km.

Exercise 2.2 — Area of Circles

Q1. The circumferences of two circles are in the ratio 5 : 6. Find the ratio of their areas.

Solution
GivenC₁ : C₂ = 5 : 6.

Step 1 — Find the ratio of radii.

Since C = 2πr, we have C₁/C₂ = r₁/r₂.

Therefore, r₁/r₂ = 5/6.

Step 2 — Find the ratio of areas.

A = πr², so A₁/A₂ = πr₁²/πr₂² = (r₁/r₂)²

= (5/6)²

= 25/36

AnswerThe ratio of their areas = 25 : 36.

Q2. Find the area of a circle whose circumference is 88 cm.

Solution
GivenCircumference C = 88 cm.

Step 1 — Find the radius from the circumference.

C = 2πr

88 = 2 × (22/7) × r

88 = (44/7) × r

r = 88 × 7 / 44

r = 2 × 7

r = 14 cm

Step 2 — Find the area.

A = πr²

= (22/7) × 14 × 14

= (22/7) × 196

= 22 × 28

= 616 cm²

AnswerArea of the circle = 616 cm².

Q3. Find the cost of polishing a circular table top of diameter 16 dm at ₹15 per dm².

Solution
GivenDiameter d = 16 dm, Cost per dm² = ₹15.

Step 1 — Find the radius.

r = d/2 = 16/2 = 8 dm

Step 2 — Find the area of the table top.

A = πr²

= (22/7) × 8 × 8

= (22/7) × 64

= 1408/7

≈ 201.14 dm²

Step 3 — Find the total cost.

Cost = ₹15 × 201.14

= ₹3,017.10

≈ ₹3,017

AnswerCost of polishing = ₹3,017 (approximately).

Q4. Find the area of a circle with radius 7 cm.

Solution
GivenRadius r = 7 cm.

A = πr²

= (22/7) × 7 × 7

= 22 × 7

= 154 cm²

AnswerArea of the circle = 154 cm².

Q5. The area of a circle is 616 cm². Find its radius.

Solution
GivenArea A = 616 cm².

Step 1 — Use the area formula to find r.

A = πr²

616 = (22/7) × r²

r² = 616 × 7 / 22

r² = 616/22 × 7

r² = 28 × 7

r² = 196

r = √196

r = 14 cm

AnswerRadius of the circle = 14 cm.

Q6. The radius of a circular park is 35 m. Find the cost of fencing it at ₹8 per metre.

Solution
GivenRadius r = 35 m, Cost per metre = ₹8.

Step 1 — Find the circumference (perimeter of the park).

C = 2πr

= 2 × (22/7) × 35

= 2 × 22 × 5

= 220 m

Step 2 — Find the total cost.

Cost = ₹8 × 220 = ₹1,760

AnswerCost of fencing = ₹1,760.

Exercise 2.3 — Combined Shapes and Applications

Q1. From a circular sheet of radius 5 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.

Solution
GivenOuter radius R = 5 cm, Inner radius r = 3 cm.

Step 1 — Find the area of the original (larger) circle.

A_outer = πR² = (22/7) × 5² = (22/7) × 25 = 550/7 cm²

Step 2 — Find the area of the removed (smaller) circle.

A_inner = πr² = (22/7) × 3² = (22/7) × 9 = 198/7 cm²

Step 3 — Find the remaining area.

Remaining area = A_outer − A_inner = π(R² − r²)

= (22/7) × (25 − 9)

= (22/7) × 16

= 352/7

≈ 50.28 cm²

AnswerArea of the remaining sheet ≈ 50.28 cm² (or exactly 352/7 cm²).

Q2. A picture is painted on cardboard 8 cm long and 5 cm wide with a 1.5 cm margin on each side. Find the total area of the margin.

Solution
GivenCardboard length = 8 cm, width = 5 cm, Margin = 1.5 cm on each side.

Step 1 — Find the area of the full cardboard.

Outer area = length × width = 8 × 5 = 40 cm²

Step 2 — Find the inner dimensions (after removing the margin from all sides).

Margin is on both left and right, so inner length = 8 − 2 × 1.5 = 8 − 3 = 5 cm

Margin is on both top and bottom, so inner width = 5 − 2 × 1.5 = 5 − 3 = 2 cm

Step 3 — Find the inner (picture) area.

Inner area = 5 × 2 = 10 cm²

Step 4 — Find the margin area.

Margin area = Outer area − Inner area = 40 − 10 = 30 cm²

AnswerTotal area of the margin = 30 cm².

Q3. A circular piece of radius 2 cm is cut from a rectangular sheet of length 5 cm and breadth 3 cm. Find the area of the remaining sheet.

Solution
GivenRectangle: l = 5 cm, b = 3 cm; Circle radius r = 2 cm.

Step 1 — Find the area of the rectangular sheet.

Rectangle area = l × b = 5 × 3 = 15 cm²

Step 2 — Find the area of the circular piece cut out.

Circle area = πr² = (22/7) × 2² = (22/7) × 4 = 88/7 ≈ 12.57 cm²

Step 3 — Find the remaining area.

Remaining area = Rectangle area − Circle area

= 15 − 88/7

= (105 − 88)/7

= 17/7

≈ 2.43 cm²

AnswerArea of the remaining sheet ≈ 2.43 cm² (or exactly 17/7 cm²).

Q4. A path of width 2 m is built around a circular park of radius 7 m. Find the area of the path.

Solution
GivenInner radius (park) r = 7 m, Path width = 2 m.

Step 1 — Find the outer radius.

Outer radius R = r + path width = 7 + 2 = 9 m

Step 2 — Find the area of the path using the ring formula.

Path area = π(R² − r²)

= (22/7) × (9² − 7²)

= (22/7) × (81 − 49)

= (22/7) × 32

= 704/7

≈ 100.57 m²

AnswerArea of the path ≈ 100.57 m² (or exactly 704/7 m²).

Q5. A wire is bent into a circle of radius 14 cm. If it is rebent into a square, find the side of the square and its area.

Solution
GivenRadius of circle r = 14 cm.

Step 1 — Find the circumference of the circle (which equals the perimeter of the square).

C = 2πr

= 2 × (22/7) × 14

= 2 × 22 × 2

= 88 cm

Step 2 — Find the side of the square.

The same wire is rebent, so Perimeter of square = Circumference of circle.

4 × side = 88

side = 88/4 = 22 cm

Step 3 — Find the area of the square.

Area = side × side = 22 × 22 = 484 cm²

AnswerSide of the square = 22 cm; Area of the square = 484 cm².

Multiple Choice Questions (MCQs)

Q1. The circumference of a circle of diameter 7 cm is:

(a) 14 cm   (b) 22 cm   (c) 44 cm   (d) 11 cm

Solution

C = πd = (22/7) × 7 = 22 cm

Answer(b) 22 cm

Q2. The area of a circle with radius 14 cm is:

(a) 308 cm²   (b) 44 cm²   (c) 616 cm²   (d) 88 cm²

Solution

A = πr² = (22/7) × 14 × 14 = 22 × 2 × 14 = 616 cm²

Answer(c) 616 cm²

Q3. If the radius of a circle is doubled, the area becomes:

(a) 2 times   (b) 3 times   (c) 8 times   (d) 4 times

Solution

Original area = πr². New radius = 2r, so new area = π(2r)² = 4πr² = 4 × original area.

Answer(d) 4 times

Q4. The perimeter of a semicircular disc of radius r is:

(a) πr   (b) πr + 2r   (c) 2πr + r   (d) πr + r

Solution

Perimeter of a semicircular disc = curved part + diameter = πr + 2r.

Answer(b) πr + 2r

Q5. The ratio of the area of a circle to its circumference (when r = 7) is:

(a) 7 : 2   (b) 2 : 7   (c) 7 : 1   (d) 1 : 7

Solution

Area = πr² = (22/7) × 49 = 154 cm²

Circumference = 2πr = 2 × (22/7) × 7 = 44 cm

Ratio = 154 : 44 = 7 : 2

Answer(a) 7 : 2

True or False

1. The circumference of a circle is always greater than its diameter.

SolutionC = πd ≈ 3.14 × d, which is always greater than d (since π > 1).
AnswerTrue

2. Area of a circle = πd².

SolutionThe correct formula is A = πr². Since r = d/2, we get A = π(d/2)² = πd²/4, not πd².
AnswerFalse — the correct formula is A = πr² or A = πd²/4.

3. If the radius doubles, the circumference also doubles.

SolutionC = 2πr. If r becomes 2r, then C = 2π(2r) = 2 × (2πr) = twice the original circumference.
AnswerTrue

4. The perimeter of a semicircle includes the diameter.

SolutionA semicircular disc has a curved part (πr) and a straight edge which is the diameter (2r). So the perimeter is πr + 2r, which includes the diameter.
AnswerTrue

5. Two circles with the same area have the same radius.

SolutionA = πr². If A₁ = A₂, then πr₁² = πr₂², which means r₁² = r₂², so r₁ = r₂ (since radius is positive).
AnswerTrue

Fill in the Blanks

1. The circumference of a circle with radius r is ___.

Answer2πr

The circumference (distance around the circle) equals 2πr, where r is the radius.

2. The area of a circle with diameter d is ___.

Answerπd²/4

Since d = 2r, we have r = d/2. Substituting: A = πr² = π(d/2)² = πd²/4.

3. If r = 7 cm, then the area of the circle is ___ cm².

Answer154

A = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm².

4. The perimeter of a semicircular disc of radius r is ___.

Answerπr + 2r

The curved part is πr (half of 2πr) and the straight edge is the diameter 2r.

5. If the radius is tripled, the area becomes ___ times.

Answer9

A = πr². New radius = 3r, so new area = π(3r)² = 9πr² = 9 times the original.

Common Mistakes

1. Confusing circumference (2πr) with area (πr²) — Always pause and ask: "Am I finding the distance around the circle (circumference) or the space inside it (area)?" before choosing the formula.

2. Forgetting to add the diameter for the perimeter of a semicircular disc — The perimeter of a semicircular disc is πr + 2r, not just πr. The straight edge (diameter) must be included.

3. Using diameter instead of radius in the area formula — A = πr² uses the radius. If the diameter is given, first convert: r = d/2, then compute A = π(d/2)² = πd²/4. Never write A = πd².

4. Not converting units consistently — A wheel diameter in cm but distance in km? Convert everything to the same unit first. 1 km = 1000 m = 1,00,000 cm.

5. Forgetting to multiply by π — Some students compute 2r instead of 2πr, or r² instead of πr². The factor of π is essential in every circle formula.

6. Confusing "area of path" with "area of whole region" — When a path surrounds a circle, the path area = π(R² − r²), which is the area of the outer circle minus the area of the inner circle. Do not just compute πR².

7. Not squaring the radius in area calculations — A = πr² means r × r, not just π × r. Writing πr instead of πr² is a very common error. Always verify: r = 7 → r² = 49, not 7.

Exam Checklist

Before you walk into the exam, make sure you can do each of these without looking:

  • [ ] Circumference formula: C = 2πr (from radius) or C = πd (from diameter)
  • [ ] Area formula: A = πr² (always from radius; if diameter given, halve it first)
  • [ ] Perimeter of a semicircular disc: πr + 2r (curved part + diameter)
  • [ ] Area between two concentric circles: π(R² − r²)
  • [ ] Word problems involving cost: Find the circumference or area first, then multiply by the cost per unit
  • [ ] Revolution problems: One revolution = one circumference; number of revolutions = total distance ÷ circumference
  • [ ] Ratio problems: Ratio of circumferences = ratio of radii; ratio of areas = ratio of squares of radii
  • [ ] Combined shapes: Rectangle minus circle, circular path around a circle, wire rebent from one shape to another
  • [ ] Unit conversions: cm ↔ m ↔ km — always convert to a single unit before computing

Related Solutions & Guides

  • [Term 2 Number System Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-2-number-system-exercise-solutions)
  • [Term 2 Algebra Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-2-algebra-exercise-solutions)
  • [All Class 7 Maths Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions)
  • [Class 7 Maths Subject Guide](/school/tamil-nadu-state-board/class/7/maths)

Frequently Asked Questions

The circumference of a circle is C = 2πr where r is the radius, or C = πd where d is the diameter. Using π = 22/7 or 3.14, you can calculate the distance around any circle.